Give n 32bit numbers which are all unsigned integers, then find the two numbers that can get the maximum value after their XOR. e.g. for these numbers

Give n 32bit numbers which are all unsigned integers, then find the two numbers that can get the maximum value after their XOR.
e.g.
for these numbers: 1, 2, 3, 4, 0xFFFFFFFE,
we should return 1 and 0xFFFFFFFE whose XOR result is the max.
please give a algorithm with time complexity <= nlgn, O(n)would be better.
I think this can be done in O(n).
I have a two-pass algorithm:

1. The first pass builds a binary tree.
. The tree has n leaves corresponding to the n given numbers.
. Each leaf has a depth of 32.
. When you add a leaf, you start from the MSB, if 0 go left, if 1 go right.
2. In the second pass, for each number x, bit complement (m=~x) it first, and then find the best "match" of m in the tree.
The "match" looks like the following:

  unsigned long BestMatch(unsigned long m, node * pRoot)  
  {
      unsigned long match = 0;
      for (int i=31; i>=0; i--)
      {
          match <<= 1;
          BYTE bit = (m>>i)&0x01;
          if (bit)
          {
              if (p->right)
              {
                  p = p->right;
                  match ++;
              }
              else
              {
                  p = p->left;
              }
          }
          else
          {
              if (p->left)
              {
                  p = p->left;
                  match ++;
              }
              else
              {
                  p = p->right;
              }
          }
      }
      return match;
  }